in the textbook by Sydsæter, Hammond, Seierstad and Strøm that the general ....
Exercise 9 Solve the equation in the case f (K) = bK, U (C) = C1-v/(1 − v). 3.
BEEM103 –Optimization Techniques for Economists
Dieter Balkenborg Departments of Economics University of Exeter
Class Exercises Week 6 - Solutions
Exercise 1 Find the solutions of the form e x•
t
2x_ + x = 0
Solution 1 For such a solution x_ = e t , x• = x• A solution is given by
2x_ + x =
to the homogenous equation
2
e
t
and hence
2 +1 e
t
=(
2
1)2 e
t
= 1, i.e. x (t) = et .
Exercise 2 Find a solution to the inhomogeneous equation x•
2x_ + x = 3
Solution 2 A solution is given by x (t) = 3. Exercise 3 Describe all solutions to this equation. Solution 3 Because 2 2 + 1 has the double root = 1 it follows from Theorem 6.3.1 in the textbook by Sydsæter, Hammond, Seierstad and Strøm that the general solution to the homogeneous equation x• 2x_ + x = 0 is (A + Bt) et . The same book discusses two pages later that the general solution to the inhomogeneous solution is (A + Bt) et + 3 Exercise 4 Find a solution with x (0) = 0, x (1) = 1 Solution 4 x (0) = A + 3 = 0 () A = 3 x (1) = ( 3 + B) e + 3 = 2 ()
3+B =
e
1
() B =
e
1
+3
Exercise 5 Find the solutions to the homogenous equation x• + x = 0 Solution 5 The characteristic equation 2 + 1 has no real solution. By Theorem 6.3.1 in the textbook by Sydsæter, Hammond, Seierstad and Strøm the general solution is x = A cos t + B sin t Exercise 6 Minimize
Z
1
p 1 + x_ 2 dt
0
subject to the boundary conditions x (0) = 4, x (1) = 2:
Solution 6 Let F =
p
1 + x_ 2 . We have
1 1 @F 1 @F = 0, = 1 + x_ 2 2 2x_ = 1 + x_ 2 2 x_ @x @ x_ 2 @ 2F @ 2F = =0 @ x@t _ @ x@x _ 3 1 @ 2F 1 = 1 + x_ 2 2 2x_ x_ + 1 + x_ 2 2 @ x@ _ x_ 2 1 x_ 2 1 1 = p +1 = p >0 2 2 2 1 + x_ 1 + x_ 1 + x_ 1 + x_ 2
The Euler equation is hence in this case 1 1 0= p x• () x• = 0 2 1 + x_ 1 + x_ 2 The general solution to x• = 0 is x = At + B. To …t the boundary conditions we must have x (0) = B = 4 x (1) = A + 4 = 2 , A = Thus x (t) =
2
2t + 4
Exercise 7 Minimize
Z
1
x2 + 2txx_ + t2 x_ 2 dt
0
subject to the boundary conditions x (0) = 0, x (1) = 2: Solution 7 @F = 2x + 2tx, _ @x @ 2F = 2x + 4tx, _ @ x@t _
@F = 2tx + 2t2 x_ @ x_ @ 2F @ 2F = 2t, = 2t2 @ x@x _ @ x@ _ x_
The Euler equation is 2x + 2tx_ 0
2x + 4tx_ + 2tx_ + 2t2 x• 4tx_ + 2t2 x• 2 () x• = x_ for t 6= 0 t = =
Let u = x. _ Then the last equation becomes du = dt
2
2 u t
This is a separable di¤erential equation which we can solve as follows 1 du Z u 1 du u ln juj ln juj juj
= = = = =
2 dt t Z 2 dt t 2 ln jtj + C ln (jtj) 2 + C = ln t t 2 or u = t 2
2
where = ln C. Next dx = u= dt Z x =
t
2
t 2 dt =
t
1
+D
The only function possible which could in addition satisfy x (0) = 0 would be obtained for = 0 (otherwise the function would not be de…ned at zero) and A = 0, i.e., we would have x (t) = 0 for all t. However, x (2) = 2 would then not be possible. Hence we cannot …nd a solution and suspect that none exists. Exercise 8 Find the Euler equation for the Ramsey problem Z T U f (K (t)) K_ (t) e rt dt 0
subject to the boundary conditions x (0) = 4, x (1) = 2: Use the relative rate of change _ _ consumption C=C where C = f (K) K. Solution 8 For this and the next exercise see Sydsæter, Hammond, Seierstad and Strøm Section 8.4: Exercise 9 Solve the equation in the case f (K) = bK; U (C) = C 1 v = (1
3
v).
